```
package com.thealgorithms.datastructures.trees;
import com.thealgorithms.datastructures.trees.BinaryTree.Node;
/**
* Problem Statement Ceil value for any number x in a collection is a number y
* which is either equal to x or the least greater number than x.
*
* Problem: Given a binary search tree containing positive integer values. Find
* ceil value for a given key in O(lg(n)) time. In case if it is not present
* return -1.
*
* Ex.1. [30,20,40,10,25,35,50] represents level order traversal of a binary
* search tree. Find ceil for 10. Answer: 20
*
* Ex.2. [30,20,40,10,25,35,50] represents level order traversal of a binary
* search tree. Find ceil for 22 Answer: 25
*
* Ex.2. [30,20,40,10,25,35,50] represents level order traversal of a binary
* search tree. Find ceil for 52 Answer: -1
*/
/**
*
* Solution 1: Brute Force Solution: Do an inorder traversal and save result
* into an array. Iterate over the array to get an element equal to or greater
* than current key. Time Complexity: O(n) Space Complexity: O(n) for auxillary
* array to save inorder representation of tree.
* <p>
* <p>
* Solution 2: Brute Force Solution: Do an inorder traversal and save result
* into an array.Since array is sorted do a binary search over the array to get
* an element equal to or greater than current key. Time Complexity: O(n) for
* traversal of tree and O(lg(n)) for binary search in array. Total = O(n) Space
* Complexity: O(n) for auxillary array to save inorder representation of tree.
* <p>
* <p>
* Solution 3: Optimal We can do a DFS search on given tree in following
* fashion. i) if root is null then return null because then ceil doesn't exist
* ii) If key is lesser than root value than ceil will be in right subtree so
* call recursively on right subtree iii) if key is greater than current root,
* then either a) the root is ceil b) ceil is in left subtree: call for left
* subtree. If left subtree returns a non null value then that will be ceil
* otherwise the root is ceil
*/
public class CeilInBinarySearchTree {
public static Node getCeil(Node root, int key) {
if (root == null) {
return null;
}
// if root value is same as key than root is the ceiling
if (root.data == key) {
return root;
}
// if root value is lesser than key then ceil must be in right subtree
if (root.data < key) {
return getCeil(root.right, key);
}
// if root value is greater than key then ceil can be in left subtree or if
// it is not in left subtree then current node will be ceil
Node result = getCeil(root.left, key);
// if result is null it means that there is no ceil in children subtrees
// and the root is the ceil otherwise the returned node is the ceil.
return result == null ? root : result;
}
}
```